» I have another problem that involves some sort of substitution.

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John E. Franklin

2005-12-16 17:47:11

oh. I'll examine tomorrow. bye.

sirsosay

2005-12-16 17:46:22

Last equation was the derivative after being simplified.

John E. Franklin

2005-12-16 17:43:51

Okay, I get the product rule. But what is that last equation you wrote? Also, the math script isn't necessary unless you like to type even more to get it to look cool. As long as your understood. But [m a t h]\frac{3}{4}[/m a t h] makes a 3/4 fraction. You can google on LaTeX Math to learn it. \int is integrate _{x=0} makes x=0 a subscript, the underscore does that. ^{I'm tiny north east} makes writing a superscript.

I'm going to bed now, it's 1:45 am here. Bye.

sirsosay

2005-12-16 17:28:15

The derivative of...

would use the product rule.

f'g+g'f

+ J*(-1/(x+1)^2)

I don't know how to use th math script..

(x^2+2x+2)/(x+1)^2

John E. Franklin

2005-12-16 17:09:25

Well I cheated after that and plotted the points with a tiny BASIC program. And noted the minimum area was 4 at x=2

I'm rusty at calculus, but I want to learn it again years later. So this is a product, so there is a rule for that. What is the derivative of this? dJ

sirsosay

2005-12-16 17:08:45

Wow thats really ingenious. Thats cool.

I'll show my teacher your method.

Ricky

2005-12-16 17:07:36

Ah, I see now. Guess no calculus is needed.

John E. Franklin

2005-12-16 17:07:28

Add 2 to both sides to solve for Yintercept.

Multiply both sides of equation by Xintercept so as to solve for the area; you can divide by 2, if you want to be exact.

John E. Franklin

2005-12-16 17:01:07

Click on picture to see it bigger.

sirsosay

2005-12-16 17:00:58

I think what he may have done is made a triangle by drawing a line from the origin to the point and making a right triangle.

What I'm most curious is if this congruency is true at all points.

Ricky

2005-12-16 16:48:59

John, I'm curious how you came up with that. Think you could show more work? I can't seem to find where that equation is coming from.

John E. Franklin

2005-12-16 16:29:04

Slope = -2

X intercept = 2 Y intercept = 4

sirsosay

2005-12-16 16:21:45

Thank you both for helping.

John E. Franklin

2005-12-16 16:20:12

The x intercept must be greater than one. Call x intercept J. Here is the equation for the area. Just find the J above one that makes the expression the smallest:

I arrived at this by drawing a picture and setting two similar triangles proportions equal. They are:

Sorry explanation so poor.

Ricky

2005-12-16 15:31:33

Start with the basics.

y = mx + b

We know this has to pass through the point (x, y) = (1, 2), so lets plug those in:

2 = m + b, or, m = 2 - b

That didn't seem to get us very far, did it? But at least we have a relationship for m and b, this might come in handy later.

So lets go back. What we want is a general equation for the area of the triangle. Well, we know of the formula 1/2 * base * height. So lets try to find those variables.

The height of the triangle is going to be the y-intercept, b. Easy enough.

The base of the triangle is going to be the x intercept. Since y = mx + b, and y must be 0 (definition of the x-intercept) 0 = mx + b. We want to solve this for x, so that would be x = -b / m.

So the area of the triangle is 1/2 * b * -b/m, or -b^2 / 2m. But wait, isn't this going to be negative? Negative area? Nope, remember the line that we are drawing has a negative slope, so m is negative, making -b^2 / m positive.

So we want to find the least area of the function -b^2 / 2m. Huh, two variables, that's going to be pretty tricky without multi variable calculus. But wait, doesn't m = 2 - b? Told you that would come in handy. So -b^2 / 2 * (2 - b) is the area, or -b^2 / (4 - 2b).

Try to find the minimum for that function. This will tell you what b is, then you can find m because m = 2 - b.

Edit:

And for extra credit, what kind of triangle does this make?